Question

inte
$integration of x squar by x square plus one$
​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \int \dfrac{x^{2} }{x^{2} + 1} dx$

As we know that,

Adding and subtracting 1 in numerator, we get.

$\sf \implies \displaystyle \int \dfrac{x^{2} + 1 - 1}{x^{2} + 1} dx$

$\sf \implies \displaystyle \int \dfrac{x^{2} + 1}{x^{2} + 1} dx \ - \int \dfrac{1}{x^{2} + 1} dx$

$\sf \implies \displaystyle \int 1.dx \ - \int \dfrac{1}{x^{2} + 1} dx$

$\sf \implies \displaystyle x - \int \frac{1}{x^{2} + 1} dx$

Substitute the value of x = tan(t) in equation, we get.

$\sf \implies \displaystyle x - \int \dfrac{1.dx}{tan^{2}t + 1 }$

Differentiate the equation w.r.t x, we get.

⇒ x = tan(t).

⇒ dx = sec²tdt.

Substitute the value in the equation, we get.

$\sf \implies \displaystyle x - \int \dfrac{sec^{2}t .dt }{tan^{2}t + 1 }$

As we know that,

Formula of :

⇒ 1 + tan²x = sec²x.

Using this formula in equation, we get.

$\sf \implies \displaystyle x - \int \dfrac{sec^{2}t.dt }{sec^{2} t}$

$\sf \implies \displaystyle x - \int dt$

$\sf \implies \displaystyle x - t + C$

⇒ x = tan(t).

⇒ t = tan⁻¹(x).

Put the value of x = tan⁻¹x in equation, we get.

$\sf \implies \displaystyle x - tan^{-1} x + C.$

$\sf \implies \displaystyle \int \dfrac{x^{2} }{x^{2} + 1} dx = x - tan^{-1} x + C.$

                                                                                                                   

MORE INFORMATION.

Standard integrals.

(1) = ∫0.dx = c.

(2) = ∫1.dx = x + c.

(3) = ∫k dx = kx + c, (k ∈ R).

(4) = ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠ - 1).

(5) = ∫dx/x = ㏒(x) + c.

(6) = ∫eˣdx = eˣ + c.

(7) = ∫aˣdx = aˣ/㏒(a) + c = aˣ㏒(e) + c.